#-(and) " P64 (**) Layout a binary tree (1) Given a binary tree as the usual Prolog term t(X,L,R) (or nil). As a preparation for drawing the tree, a layout algorithm is required to determine the position of each node in a rectangular grid. Several layout methods are conceivable, one of them is shown in the illustration below. [p64] In this layout strategy, the position of a node v is obtained by the following two rules: □ x(v) is equal to the position of the node v in the inorder sequence □ y(v) is equal to the depth of the node v in the tree In order to store the position of the nodes, we extend the Prolog term representing a node (and its successors) as follows: % nil represents the empty tree (as usual) % t(W,X,Y,L,R) represents a (non-empty) binary tree with root W % \"positioned\" at (X,Y), and subtrees L and R Write a predicate layout-binary-tree/2 with the following specification: % layout-binary-tree(T,PT) :- PT is the \"positioned\" binary tree % obtained from the binary tree T. (+,?) Test your predicate in an appropriate way. " (load "p54a") ;; To add the coordinate, we create a new structure, which inherits ;; from the binary-tree structure, so we can reuse that abstraction. ;; However, including structures will make the new fields added at the ;; end of it. The order of the fields should be immaterial (only that ;; we don't use true structures, but lists, so the new fields are ;; added at the end of the lists, compared to the included list ;; structures). (defstruct (layout-binary-tree (:include binary-tree) (:type list)) x y) (defun binary-tree-to-layout-binary-tree (tree) " Return a layout-binary-tree homologue to node. " (if (binary-tree-empty-p tree) (make-empty-binary-tree) (make-layout-binary-tree :label (binary-tree-label tree) :left (binary-tree-to-layout-binary-tree (binary-tree-left tree)) :right (binary-tree-to-layout-binary-tree (binary-tree-right tree))))) ;; To layout the binary tree, we will do it in two steps. First we ;; make the layout tree, and setting the y field to the depth of each ;; node. Then we execute a infix walk of the new tree updating the x ;; field of each node. (defun layout-node-depth (node depth) " Return a layout-binary-tree homologue to node, with the ordinates of each node set to their depth. " (if (binary-tree-empty-p node) (make-empty-binary-tree) (make-layout-binary-tree :label (binary-tree-label node) :y depth :left (layout-node-depth (binary-tree-left node) (1+ depth)) :right (layout-node-depth (binary-tree-right node) (1+ depth))))) ;; Note, incf is a prefix increment, it returns the new-value. ;; Therefore it is easier to start with the predecessor of the first ;; value, and to finally return the last value used. One could define ;; a postfix increment operator to easily write the code using the ;; other convention. (defun layout-node-abscissa/inorder (node abscissa) " Sets the abscissa of each node in the subtree NODE to a sequence of values starting from (1+ ABSCISSA) for the left-most node. Returns the last abscissa used. " (cond ((binary-tree-empty-p (binary-tree-left node)) (setf (layout-binary-tree-x node) (incf abscissa))) ((binary-tree-empty-p (binary-tree-right node)) (setf (layout-binary-tree-x node) (layout-node-abscissa/inorder (binary-tree-left node) abscissa))) (t (layout-node-abscissa/inorder (binary-tree-right node) (1+ (setf (layout-binary-tree-x node) (layout-node-abscissa/inorder (binary-tree-left node) abscissa))))))) (defun layout-binary-tree/inorder/depth (tree) "" (let ((lobt (layout-node-depth tree 0))) (layout-node-abscissa/inorder lobt 0) ; starts from 1; use -1 to start from 0. lobt)) ;; (layout-binary-tree/inorder/depth (complete-binary-tree 5)) ;;;; THE END ;;;;